3 The Quadratic Function A quadratic function can be expressed in three formats:f(x) = ax2 + bx + c is called the standard formis called the factored formf (x) = a(x - h)2 + k, is called the vertex form
4 The Standard Form f(x)=ax2 + bx + c If a > 0 then parabola opens up If a < 0 then parabola opens down(0,c) is the y-interceptAxis of symmetry is
5 The Vertex Form f (x) = a(x - h)2 + k, a 0 ***y – k = a(x-h)2The graph of f is a parabola whose vertex is the point (h, k)The parabola is symmetric to the line x = hIf a > 0, the parabola opens upwardif a < 0, the parabola opens downward
6 Graphing Parabolas-Vertex Form To graph f (x) = a(x - h)2 + k:Determine whether the parabola opens up or down.If a > 0, it opens up.If a < 0, it opens down.Determine the vertex of the parabola--- V(h, k).Find any x-intercepts.Replace f (x) with 0.Solve the resulting quadratic equation for x.Find the y-intercept by replacing x with zero.Plot the intercepts and vertex. Connect these points with a smooth curve that is shaped like a cup.
7 The Factored Form Factored Form of a quadratic r represents roots or x interceptsAxis of symmetry:
8 Vertex form f (x) = a(x - h)2 + k ExampleGraph the quadratic function f (x) = -2(x - 3)2 + 8.Vertex form f (x) = a(x - h)2 + k a=-2 h =3 k = 8Given equation f (x) = -2(x - 3)2 + 8This is a parabola that opens down with vertex V(3,8)
9 Example cont. The x-intercepts are 1 and 5. Find the vertex. The vertex of the parabola is at (h, k). Because h = 3 and k = 8, the parabola has its vertex at (3, 8).Find the x-intercepts. Replace f (x) with 0: f (x) = -2(x - 3)2 + 8.0 = -2(x - 3) Find x-intercepts, setting f (x) equal to zero.2(x - 3)2 = 8(x - 3)2 = Divide both sides by 2.(x - 3) = ± Apply the square root method.x - 3 = or x - 3 = Express as two separate equations.x = or x = Add 3 to both sides in each equation.The x-intercepts are 1 and 5.The parabola passes through (1, 0) and (5, 0).
10 Example cont. Graph the parabola. With a vertex at (3, 8) f (0) = -2(0 - 3) = -2(-3) = -2(9) + 8 = -10Find the y-intercept. Replace x with 0 in f (x) = -2(x - 3)2 + 8.The y-intercept is –10. The parabola passes through (0, -10).Graph the parabola.With a vertex at (3, 8)x-intercepts at (1,0) and (5,0)and a y-intercept at (0,–10)the axis of symmetry is the vertical line x = 3.
11 The Vertex of a Parabola Whose Equation Is f (x) = ax 2 + bx + c Consider the parabola defined by the quadratic functionf (x) = ax 2 + bx + c. The parabola's vertex is at
12 Example Graph the quadratic function f (x) = -x2 + 6x -. Determine how the parabola opens.Note that a, the coefficient of x 2, is -1. Thus, a < 0;This negative value tells us that the parabola opens downward.Find the vertex.x = -b/(2a).Identify a, b, and c to substitute the values into the equation for the x-coordinate:x = -b/(2a) = -6/2(-1)=3.The x-coordinate of the vertex is 3.We substitute 3 for x in the equation of the function to find the y-coordinate:y=f(3) = -(3)^2+6(3)-2= =7,the parabola has its vertex at (3,7).
13 Example Graph the quadratic function f (x) = -x2 + 6x -. Find the x-intercepts. Replace f (x) with 0 in f (x) = -x2 + 6x - 2.0 = -x2 + 6x (If you cannot factor)
14 Example Graph the quadratic function f (x) = -x2 + 6x -. Find the y-intercept. Replace x with 0 in f (x) = -x2 + 6x - 2.f (0) = • = -The y-intercept is –2. The parabola passes through (0, -2).Graph the parabola.
15 Minimum and Maximum: Quadratic Functions Consider f(x) = ax2 + bx +c.If a > 0, then f has a minimum that occurs at x = -b/(2a). This minimum value is f(-b/(2a)).If a < 0, the f has a maximum that occurs at x = -b/(2a). This maximum value is f(-b/(2a)).
16 The Discriminant and the Kinds of Solutions to ax2 + bx +c = 0 No x-interceptsNo real solution;two complex imaginary solutionsb2 – 4ac < 0One x-interceptOne real solution(a repeated solution)b2 – 4ac = 0Two x-interceptsTwo unequal real solutionsb2 – 4ac > 0Graph ofy = ax2 + bx + cKinds of solutionsto ax2 + bx + c = 0Discriminantb2 – 4ac